ro1=ms1/Vs1=>ms1=400*1.2=480g H2SO4
Cp1=md1/ms1*100=>md1=28*480/100=134.4g H2SO4
ro2=ms2/Vs2=>ms2=1.17*620=725.4g H2So4
Cp2=md2/ms2*100=>md2=24*725.4/100=174.096g H2SO4
1.5dm3=1500cm3
ro3=ms3/Vs3=>ms3=1*1500=1500g BaCl2
Cp3=md3/ms3*100=>md3=0.05*1500/100=0.75g BaCl2
mdf H2SO4 = md1+md2=134.4+174.096=308.496g H2SO4
mdf BaCl2=0.75g
n=mdf/M=308.496/98=3.1479 moli H2SO4
n=mdf/M=0.75/208=0.0036 moli BaCl2
m acid în exces => 3.1479 > 0.0036
H2SO4 + BaCl2 = BaSO4 + 2HCl
98g H2SO4.........208g BaCl2
x=0.35336g...........0.75g BaCl2
m nereactionat => 308.496-0.35336=308,142g H2SO4
Nu sunt sigur dacă asta este rezolvarea..